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24 June, 03:41

A cosmic ray electron moves at 6.5 x 106 m/s perpendicular to the Earth's magnetic field at an altitude where field strength is 1.24 10-5 T. What is the radius (in meters) of the circular path the electron follows? The charge of an electron is 1.6 x 10-19 C and its mass is 9.11 x 10-31m. Your answer should be a number with two decimal places, do not include the unit.

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  1. 24 June, 04:23
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    Answer: R=2.98 m

    Explanation: In order to solve this answer we have to use the dynamic of circular movement actioned by the Lorentz force, this is given by:

    Fm=qvB (condiring that v and B are perpendicular)

    By using the second Newton law fro a circular movement, we have:

    Fm=m*a=m*v^2/R where v and R are the speed and R of the circular movement electrons in the magnetic field.

    then

    qvB=m*v^2/R

    Finally R=m*v / (qB) = 9.11*10^-31*6.5*10^6 / (1.6*10^-19*1.24*10^-5) = 2.98
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